Problem: What is the area of the region between $2$ consecutive points where the graphs of $f(x)=\cos(x)$ and $g(x)=-\cos(x)+2$ intersect? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $4$ (Choice C) C $2\pi$ (Choice D) D $4\pi$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${2}$ ${4}$ ${6}$ ${1}$ ${2}$ ${3}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( g(x)-f(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of two consecutive points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ \cos(x)&=-\cos(x)+2\\\\ 2\cos(x) &= 2 \\\\ \cos(x) &=1 \end{aligned}$ The graphs intersect when $x=0\pm 2k\pi$, where $k$ is any integer. Since we want $2$ consecutive values, let's use $k=0$ and $k=1$, which means we will use the points $x=0$ and $x=2\pi$. Note: The area between the graphs is the same between any $2$ consecutive points where they meet, so it does not matter which values we picked for $k$, as long as they were consecutive. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{0}^{2\pi}\left(-\cos(x)+2-(\cos(x))\right)\,dx \\\\ &= \int_{0}^{2\pi}\left(-2\cos(x)+2\right)\,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{2\pi}\left(-2\cos(x)+2\right)\,dx \\\\ &= -2\sin(x)+2x~\Bigg|_{0}^{2\pi} \\\\ &= \left( 0+4\pi \right) - \left( 0+0 \right) \\\\ &= 4\pi \end{aligned}$ Answer The area is $4\pi$ square units.